# Test Latex Logika

$$
\documentclass{article}
\newcommand{\logicarg}[2]{% \logicarg{<premise>}{<conclusion>}
\begin{tabular}[t]{@{}l@{}}
#1 \\ \hline #2
\end{tabular}%
}
\begin{document}
\begin{enumerate}
\item \logicarg
{If you are eating soup then I am happy. \\ You are eating soup}
{I am happy}
\item \logicarg
{If it is Wednesday then you aren't taking a quiz. \\ It is not Wednesday.}
{You are taking a quiz.}
\item \logicarg
{Either you love logic or you hate it. \\ You don't hate logic.}
{You love logic.}
\end{enumerate}
\end{document}
$$

$$
\begin{tabular}{ l r }
p \rightarrow q \\
p \\[5pt]
\hline
\therefore \ q
\end{tabular}
$$

$$
\begin{table}[t]
\hline
&Treatment A&Treatment B\\
\hline
\end{table}
$$

$$
\begin{table}[]
\begin{tabular}{ll}
a & b \\
c & d \\
\end{tabular}
\end{table}
$$

$$
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
Johnson Family & Transformation & Parameter Conditions & X Condition \\ \hline
$S_B$ & $Z=\gamma + \eta ln(\frac {X - \epsilon} {\lambda + \epsilon - X})$ & $\eta, \lambda >0, -\infty < \gamma, \epsilon < \infty$ & $\epsilon < X < \epsilon + \lambda$ \\ \hline
$S_L$ & $Z=\gamma + \eta ln(X - \epsilon)$ & $\eta >0, -\infty < \gamma, \epsilon < \infty$ & $X > \epsilon$ \\ \hline
$S_U$ & $Z=\gamma + \eta \sinh^{-1}(\frac {X - \epsilon} {\lambda})$ & $\eta, \lambda >0, -\infty < \gamma, \epsilon < \infty$ & $-\infty < X < \infty$ \\ \hline
\end{tabular}
\end{center}
$$